1. CMPT 258 Database Systems
Final Exam Review

2. Final Exam Final exam Cheat sheet Tuesday Dec 15, 11 – 1 RLC 105
Both sides of an 8x11 paper

3. Coverage ER diagram Database design Relational algebra SQL queries
Create Table
Update, delete, etc.
Relational algebra
SQL queries
Normal forms

4. Best ways to study Lecture notes Homework problems 2 Tests
Review the examples we did in class
Homework problems
2 Tests

5. Key Constraints 1:1 1:N N:1 M:N
Many-to-Many
1-to-1
1-to Many
Many-to-1
Consider Works_In: An employee can work in many departments; a dept can have many employees.
1:1
1:N
N:1
M:N
did
dname
budget
Departments
lot
since
name
Works_In
Employees
ssn M N

6. Session #, Speaker Name
08/30/11
Key Constraints
The constraint that each department has at most one manager is an example of a key constraint, and it implies that each Departments entity appear in at most one Manages relationship.
Arrow states that given a Department entity, we can uniquely determine the manages relationship in which it appears.
dname
budget
did
since
lot
name
ssn
Manages
Employees
Departments
Many-to-Many
1-to-1
1-to Many
Many-to-1
Session #, Speaker Name
08/30/11 6 6

7. Participation Constraints
Does every department have a manager?
If so, this is a participation constraint: the participation of Departments in Manages is said to be total (vs. partial).
Every Departments entity must appear in an instance of the Manages relationship.
lot
name
dname
budget
did
since
Manages
Departments
Employees
ssn
Works_In

8. Weak Entities
A weak entity can be identified uniquely only by considering the primary key of another (owner) entity.
Partial key: the set of attributes of a weak entity set that uniquely identify a weak entity for a given owner entity
lot
name
age
pname
Dependents
Employees
ssn
Policy
cost 1 N

9. Weak Entities
Owner entity set and weak entity set must participate in a one-to-one or one-to-many relationship set (one owner, one or more weak entities).
Weak entity set must have total participation in this identifying relationship set.
identifying owner, strong entity
lot
name
age
pname
Dependents
Employees
ssn
Policy
cost

10. Aggregation
Aggregation vs. ternary relationship: why not make Sponsors a ternary relationship?
There are really two distinct relationships (monitors and sponsors), each with attribute of its own.
Monitors is a distinct relationship, with a descriptive attribute.
budget
did
pid
started_on
pbudget
dname
until
Departments
Projects
Sponsors
Employees
Monitors
lot
name
ssn
since

11. Creating Tables in SQL Creates the Students relation.
4/26/2017
Creating Tables in SQL
Creates the Students relation.
Observe that the type (domain) of each field is specified, and enforced by the DBMS whenever tuples are added or modified.
As another example, the Enrolled table holds information about courses that students take.
CREATE TABLE Students
(sid CHAR(20),
name CHAR(20),
login CHAR(10),
age INTEGER,
gpa REAL)
CREATE TABLE Enrolled
(sid CHAR(20),
cid CHAR(20),
grade CHAR(2))
Session #, Speaker Name 15

12. Altering Tables
ALTER TABLE tbl_name
ADD COLUMN col_name domain
DROP COLUMN col_name
e.g., ALTER TABLE Students
ADD COLUMN firstYear integer
The schema of Students is altered by adding a new field; every tuple in the current instance is extended with a null value in the new field.

13. DML: updating data within a table
UPDATE table_name SET column_name1 = value1, column_name2 = value2, [WHERE condition]

14. DML: Deleting Tuples
Can delete all tuples satisfying some condition (e.g., name = Smith):
DELETE FROM Students
WHERE name = ‘Smith’
What if nothing satisfies the condition?

15. Destroying Tables DROP TABLE tbl_name;
e.g., DROP TABLE Students;
e.g., DROP TABLE Enrolled;
Destroys the relation Students. The schema information and the tuples are deleted.

16. 4/26/2017
Foreign Keys in SQL
Only students listed in the Students relation should be allowed to enroll for courses.
CREATE TABLE Enrolled
(sid CHAR(10), cid CHAR(30), grade CHAR(2),
PRIMARY KEY (sid,cid),
FOREIGN KEY (sid) REFERENCES Students(sid) );
Enrolled
Students
Session #, Speaker Name 7

17. Referential Integrity in SQL
support all 4 options on deletes and updates.
Default is NO ACTION
delete/update is rejected
CASCADE
also delete all tuples that refer to deleted tuple
SET DEFAULT
Set E.sid to S.sid of the “default” student if a student is deleted from the Student relation
SET NULL
Sets foreign key value of referencing tuple to null

18. Referential Integrity in SQL
CREATE TABLE Enrolled
(sid CHAR(20),
cid CHAR(20),
grade CHAR(2),
PRIMARY KEY (sid,cid),
FOREIGN KEY (sid) REFERENCES Students(sid)
ON DELETE NO ACTION
ON UPDATE CASCADE );

19. Translating ER Diagrams with Key Constraints
4/26/2017
Translating ER Diagrams with Key Constraints
CREATE TABLE Dept_Mgr(
did INTEGER,
dname CHAR(20),
budget REAL,
ssn CHAR(11),
since DATE,
PRIMARY KEY (did),
FOREIGN KEY (ssn) REFERENCES Employees)
Since each department has a unique manager, we could instead combine Manages and Departments.
dname
budget
did
since
lot
name
ssn
Manages
Employees
Departments
Session #, Speaker Name 7

20. Review: Participation Constraints
4/26/2017
Review: Participation Constraints
Does every department have a manager?
If so, this is a participation constraint
Every did value in Departments table must appear in a row of the Manages table (with a non-null ssn value!)
CREATE TABLE Dept_Mgr(
did INTEGER,
dname CHAR(20),
budget REAL,
ssn CHAR(11) NOT NULL,
since DATE,
PRIMARY KEY (did),
FOREIGN KEY (ssn) REFERENCES Employees)
lot
name
dname
budget
did
since
Manages
Departments
Employees
ssn
Session #, Speaker Name 8

21. Translating Weak Entity Sets
4/26/2017
Translating Weak Entity Sets
Weak entity set and identifying relationship set are translated into a single table.
When the owner entity is deleted, all owned weak entities must also be deleted.
CREATE TABLE Dep_Policy (
pname CHAR(20),
age INTEGER,
cost REAL,
ssn CHAR(11) NOT NULL,
PRIMARY KEY (pname, ssn),
FOREIGN KEY (ssn) REFERENCES Employees (ssn)
ON DELETE CASCADE)
lot
name
age
pname
Dependents
Employees
ssn
Policy
cost
Session #, Speaker Name 11

22. E/R to Relations E/R diagram E Relational schema, e.g.
account=(bname, acct_no, bal) E
E = ( a1, …, an )
a1 … an R1 E2 E1
R1= ( a1, b1, c1, …, ck )
a1 …. an
c1 …. ck
b1 …. bm

23. More on relationships What about: Could have :
put b1 as the key for R1, it is also the key for E2=(b1, …., bn)
Usual strategy:
ignore R1
Add a1, c1, …., ck to E2 instead, i.e.
E2=(b1, …., bn, a1, c1, …, ck) R1 E2 E1
a1 …. an
c1 …. ck
b1 …. bm
R1= ( a1, b1, c1, …, ck )

24. More ? ? a1 …. an c1 …. ck b1 …. bm Treat as N:1 or 1:N
E2 = ( b1, …, bm ) R1
R1 = ( a1, b1, c1 …, ck )
E1 = ( a1, …, an ) R1
E2 = ( b1, …, bm , a1, c1, …, ck)
E1 = ( a1, …, an , b1, c1, …, ck) R1
E2 = ( b1, …, bm ,) R1
Treat as N:1 or 1:N
Foreign key also needs to be marked
as a candidate key

25. E/R to Relational Weak entity sets a1 …. an b1 …. bm
E1 = ( a1, …, an ) IR E2 E1
E2 = (a1, b1, …, bm )
a1 …. an
b1 …. bm

26. E/R to Relational Aggregation E1, R1, E2, E3 as before
R2 = (c1, a1, b1, d1, …, dj)
a1 …. an
b1 …. bm d1 … dj R2 E3
c1 …. ck

27. Relational Algebra: 5 Basic Operations
Selection ( s ) Selects a subset of rows from relation (horizontal).
Projection ( p ) Retains only wanted columns from relation (vertical).
Cross-product (  ) Allows us to combine two relations.
Set-difference ( — ) Tuples in r1, but not in r2.
Union (  ) Tuples in r1 and/or in r2.
Since each operation returns a relation, operations can be composed!

28. Projection S2

29. Selection () Selects rows that satisfy selection condition.
Result is a relation.
Schema of result is same as that of the input relation.

30. Union S1 S2

31. Set Difference Set-difference ( r1—r2 ): Tuples in r1, but not in r2.
S2 – S1 S2

32. Cross Product Example R1 S1
S1 X R1 =

33. Intersection S1 S2

34. Natural Join Example R1 S1
S R1 =

35. Compound Operator: Division
Useful for expressing “for all” queries like: Find sids of sailors who have reserved all boats.
For A/B attributes of B are subset of attrs of A.
May need to “project” to make this happen.
Reserves
Boats

36. Examples of Division A/B

37. See lecture notes for more examples!

38. Conceptual Evaluation Strategy
4/26/2017
Conceptual Evaluation Strategy
SELECT [DISTINCT] target-list
FROM relation-list
WHERE qualification
Semantics of an SQL query defined in terms of the following conceptual evaluation strategy:
Compute the cross-product of relation-list.
Discard resulting tuples if they fail qualifications.
Delete attributes that are not in target-list.
If DISTINCT is specified, eliminate duplicate rows.
Session #, Speaker Name

39. Expressions and Strings
4/26/2017
Expressions and Strings
Find ages of sailors whose names begin and end with B and contain at least three characters.
LIKE is used for string matching. ‘_’ stands for any one character and ‘%’ stands for 0 or more arbitrary characters (including blanks).
‘_AB%’
Comparison operators (=, <, >, etc.) can be used for string comparison
Session #, Speaker Name

40. Find sid’s of sailors who’ve reserved a red or a green boat
4/26/2017
Find sid’s of sailors who’ve reserved a red or a green boat
SELECT R.sid
FROM Boats B, Reserves R
WHERE R.bid=B.bid
AND (B.color=‘red’ OR B.color=‘green’)
UNION: Can be used to compute the union of any two union- compatible sets of tuples (which are themselves the result of SQL queries).
If we replace OR by AND in the first version, what do we get?
SELECT R.sid
FROM Boats B, Reserves R
WHERE R.bid=B.bid
AND B.color=‘red’
UNION
AND B.color=‘green’
Session #, Speaker Name

41. Nested Queries Find names of sailors who’ve reserved boat #103:
4/26/2017
Nested Queries
Find names of sailors who’ve reserved boat #103:
SELECT sname
FROM Sailors
WHERE sid IN (SELECT sid
FROM Reserves
WHERE bid=103)
Understand semantics of nested queries
For each Sailors tuple, check the qualification by computing the subquery.
Session #, Speaker Name

42. Nested Queries Find names of sailors who’ve NOT reserved boat #103:
4/26/2017
Nested Queries
Find names of sailors who’ve NOT reserved boat #103:
SELECT sname
FROM Sailors
WHERE sid NOT IN (SELECT sid
FROM Reserves
WHERE bid=103)
A very powerful feature of SQL: a WHERE clause can itself contain an SQL query!
Session #, Speaker Name

43. Nested Queries with Correlation
4/26/2017
Nested Queries with Correlation
Q: Find names of sailors who’ve reserved boat #103:
SELECT S.sname
FROM Sailors S
WHERE EXISTS (SELECT *
FROM Reserves R
WHERE R.bid=103 AND S.sid=R.sid)
Reserves
Sailors
Session #, Speaker Name

44. Nested Queries with Correlation
4/26/2017
Nested Queries with Correlation
Q: Find names of sailors who’ve not reserved boat #103:
SELECT S.sname
FROM Sailors S
WHERE NOT EXISTS (SELECT *
FROM Reserves R
WHERE R.bid=103 AND S.sid=R.sid)
EXISTS is another set comparison operator, like IN.
NOT EXISTS
Reserves
Sailors
Session #, Speaker Name

45. Division in SQL Q: Find names of sailors who’ve reserved all boats.
4/26/2017
Division in SQL
Q: Find names of sailors who’ve reserved all boats.
SELECT S.sname
FROM Sailors S
WHERE NOT EXISTS
(SELECT bid
FROM Boats
WHERE bid NOT IN
(SELECT R.bid
FROM Reserves R
WHERE R.sid=S.sid))
For each sailor we check that
there is no boat that has not been reserved by this sailor
Session #, Speaker Name

46. Aggregate Operators Significant extension of relational algebra.
4/26/2017
Aggregate Operators
Significant extension of relational algebra.
COUNT ( [DISTINCT] A)
SUM ( [DISTINCT] A)
AVG ( [DISTINCT] A)
MAX (A)
MIN (A)
single column
Q: Find the average age of all sailors
SELECT AVG (age)
FROM Sailors
Q: Find the average age of sailors with a rating of 10
Session #, Speaker Name

47. Queries with GROUP BY and HAVING
4/26/2017
Queries with GROUP BY and HAVING
SELECT [DISTINCT] target-list
FROM relation-list
WHERE qualification
GROUP BY grouping-list
HAVING group-qualification
The target-list contains (i) attribute names (ii) terms with aggregate operations (e.g., MIN (S.age)).
The attribute list (i) must be a subset of grouping- list.
Session #, Speaker Name

48. Important!
Only columns that appear in the GROUP BY clause can appear in the HAVING clause, unless they appear with an aggregate operator (COUNT, AVG, MIN, etc.)

49. 4/26/2017
Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors
Sailors instance:
SELECT S.rating, MIN (S.age)
AS minage
FROM Sailors S
WHERE S.age >= 18
GROUP BY S.rating
HAVING COUNT (*) > 1
Answer relation:
Session #, Speaker Name

50. See lecture notes for more examples!

51. Attribute Closure Compute attribute closure of X (denoted X+).
X+ = Set of all attributes A such that X  A
X+ := X
Repeat until no change: {
if there is an fd U  V in F such that U is in X+, then
add V to X+ }
Approach can also be used to find the keys of a relation.

52. Attribute Closure (example)
R = {A, B, C, D, E}
F = { B CD, D  E, B  A, E  C, AD B }
Can B infer E ?
B+ = B
B+ = BCD
B+ = BCDA
B+ = BCDAE … Yes! and B is a key for R too!
Is D a key for R?
D+ = D
D+ = DE
D+ = DEC
… Nope!

53. Normal Forms Four most commonly used normal forms are
First (1NF) normal forms
Second (2NF) normal forms
Third (3NF) normal forms
Boyce-Codd (BCNF) normal forms
Based on functional dependencies among the attributes of a relation.

54. Second Normal Forms (2NF)
Table is in 2NF when it is in 1NF and contains no partial dependencies.
Partial dependency: there is a functional dependence in which a non-key field is dependent on a subset of a candidate key.

55. Third Normal Forms (3NF)
No non-key field depends on another.
All non-key fields depend only on a candidate key.
Table is in 3NF when it is in 2NF and contains no transitive dependencies
Transitive dependency: there are functional dependencies such that X → Y, Y → Z, and X is the primary key

56. Third Normal Forms (3NF)
Table is in 3NF when it is in 2NF and contains no transitive dependencies

57. Boyce-Codd Normal Form (BCNF)
Reln R with FDs F is in BCNF if, for all X A in
A X (called a trivial FD), or
X contains a key for R.
In other words, R is in BCNF if the only non-trivial FDs that hold over R are key constraints.
Every determinant in table is a key.
Has same characteristics as primary key, but for some reason, not chosen to be primary key 23